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2 years ago

8

Caroline knows the height and the required volume of a cone-shaped vase she’s designing. Which formula can she use to determine

the radius of the vase?

1 answer:

timama [110]2 years ago

4 0

Answer:

Option B. $r=\sqrt{\frac{3V}{\pi h}}$

Step-by-step explanation:

we know that

The volume of a cone is equal to

$V=\frac{1}{3}\pi r^{2} h$

Solve for the radius r

That means-----> isolate the variable r

Multiply by 3 both sides

$3V=\pi r^{2} h$

Divide by $(\pi h)$ both sides

$r^{2}=\frac{3V}{\pi h}$

square root both sides

$r=\sqrt{\frac{3V}{\pi h}}$

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c

There is insufficient evidence to conclude that graduate students score higher, on average, on the HPI than undergraduate students

Step-by-step explanation:

From the question we are told that

The population size is $n= 650$

The sample size for graduates is $n_1 = 300$

The sample mean for graduates is $\= x _1 = 148$

The sample standard deviation for graduates is $\sigma_1 = 16$

The sample size for under-graduates is $n _2 = 350$

The sample mean for under-graduates is $\= x _2 = 153$

The sample standard deviation for graduates is $\sigma_2 = 21$

The Null hypothesis represented as

$H_0: \mu_1 - \mu_2 = 0$

The Alternative hypothesis represented as

$H_a: \mu_1 - \mu_2 < 0$

Where $\mu_1 \ and \ \mu_2$ are the population mean

Now the test statistic is mathematically represented as

$t = \frac{(\= x_1 - \= x_2 ) }{ \sqrt{ \frac{ (n_1 - 1 )\sigma_1 ^2 + (n_2 - 1)\sigma_2^2}{n_1 +n_2 -2} } * \sqrt{ \frac{1}{n_1} + \frac{1}{n_2} } }$

substituting values

$t = \frac{(148 - 153 ) }{ \sqrt{ \frac{ (300- 1 )16 ^2 + (350 - 1) 21^2}{300 +350 -2} } * \sqrt{ \frac{1}{300} + \frac{1}{350} } }$

$t = -3.37$

The p-values is mathematically evaluated as

p-value = P(Z < -3.37 ) = 0.000376

The above answer is gotten using a p-value calculator at (0.05) level of significance

Looking the p-value we see that it is less than the level of significance (0.05) so Null hypothesis is rejected

Hence there is insufficient evidence to conclude that graduate students score higher, on average, on the HPI than undergraduate students

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